Solutions Class 12th Chemistry Part I CBSE Solution

Class 12th Chemistry Part I CBSE Solution
Intext Questions Pg-37
  1. Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride(CCl4) if 22 g of…
  2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon…
  3. Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in…
  4. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous…
  5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20%…
Intext Questions Pg-41
  1. H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the…
  2. Henry’s law constant for CO2 in water is 1.67x10^8 Pa at 298 K. Calculate the quantity of…
Intext Questions Pg-47
  1. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K.…
Intext Questions Pg-55
  1. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved…
  2. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g…
  3. Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic…
  4. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0…
Exercises
  1. Define the term solution. How many types of solutions are formed? Write briefly about each…
  2. Give an example of a solid solution in which the solute is a gas.…
  3. Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass…
  4. Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous…
  5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole…
  6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and…
  7. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.…
  8. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of…
  9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl3)…
  10. What role does the molecular interaction play in a solution of alcohol and water?…
  11. Why do gases always tend to be less soluble in liquids as the temperature is raised?…
  12. State Henry’s law and mention some important applications?
  13. The partial pressure of ethane over a solution containing 6.56 × 10-3 g of ethane is 1…
  14. What is meant by positive and negative deviations from Raoult's law and how is the sign of…
  15. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal…
  16. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two…
  17. The vapour pressure of water is 12.3kPa at 300 K. Calculate vapour pressure of 1 molal…
  18. Calculate the mass of a non-volatile solute (molar mass 40 g) which should be dissolved in…
  19. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour…
  20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the…
  21. Two elements A and B form compounds having formula AB2 and AB4 . When dissolved in 20 g of…
  22. At 300 K, 36 g of glucose present in a litre of its solution has a osmotic pressure of…
  23. Suggest the most important type of intermolecular attractive interaction in the following…
  24. Based on solute-solvent interactions, arrange the following in order of increasing…
  25. Amongst the following compounds, identify which are insoluble, partially soluble and…
  26. If the density of some lake water is 1.25g/ml and contains 92 g of Na+ ions per kg of…
  27. If the solubility product of CuS is 6 × 10-16, calculate the maximum molarity of CuS in…
  28. Calculate the mass percentage of aspirin(C9H8O4) in acetonitrile(CH3CN) when 6.5 g of…
  29. Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in…
  30. Calculate the amount of benzoic acid required for preparing 250mL of 0.15M solution in…
  31. The depression in freezing point of water observed for the same amount of acetic acid,…
  32. Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is…
  33. 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of…
  34. Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water…
  35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10^5 mm Hg.…
  36. 100 g of liquid A (molar mass 140 g mol-1) was dissolved in 1000 g of liquid B (molar mass…
  37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg…
  38. Benzene and naphthalene form ideal solution over the entire range of composition. The…
  39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen…
  40. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its…
  41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2…

Intext Questions Pg-37
Question 1.

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride(CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.


Answer:

Given,


Benzene (C6H6) is 22g in the solution.


Carbon tetrachloride (CCl4) is 122g in solution


We know,


Mass Percentage of solute = 


Total mass of the solution = 22g (benzene) + 122g (carbon tetrachloride)


= 144g


Mass percentage of benzene (C6H6 = 15.28%


Note: The second method to calculate the Mass percentage of carbon tetrachloride = 100 – Mass % of C6H6


= 100 - 15.28%


= 84.72%



Question 2.

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.


Answer:

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 - 30)g = 70 g

Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol- 1

= 78 g mol- 1

∴ Number of moles of C6H6 =30/78 mol

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5

= 154 g mol- 1

∴ Number of moles of CCl4= 70/154 mol

= 0.4545 mol

Thus, the mole fraction of C6H6is given as:

= 0.3846 / (0.3846 +0.4545)

= 0.458


Question 3.

Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to500 mL.


Answer:

Molarity is the concentration of a solution expressed as the number of moles of solute per litre of solution.


Molarity = 


(a) Given,


In 4.3 L of solution there is 30 g of Co(NO3)2. 6H2O


Molar mass of Co(NO3)2.6H2O = (1 × 59 + 2 × (1 × 14 + 3 × 16) + 6 × 18)


= 291 g/mol.


∴ Moles  =  = 0.103 mol.


Now, Molarity = 


= 0.023 M


(b) Given,


30 mL of 0.5 M H2SO4 diluted to500 mL.

Applying the formula,

M1V1 = M2V2

where, M1 = 0.5 M
V1 = 30 ml
M2 = ?
V2 = 500 ml

⇒0.5M × 30 ml = M2 × 500 ml

so,M2= 0.03 M.


Question 4.

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.


Answer:

Given,

molal solution is a solution that contains 1 molecular weight of solute in a kilogram of solvent.

2.5 kg of 0.25 molal aqueous solution.


Molar mass of urea (NH2CONH2) = (2 (1 × 14 + 2 × 1) + 1 × 12 + 1 × 16)


= 60 g/mol


1000 g of water contains 0.25 mol = (0.25 × 60) g of urea.


= 15 g of urea.


Means, 1015 g of solution contains 15 g of urea


Therefore,


2500 g of solution contains = 


= 36.95 g


Hence, mass of urea required is 37 g (approx).


Question 5.

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.


Answer:

(a) Molality, also called molal concentration, is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent.


Molar mass of KI  39 + 127 = 166 g/mol.


20% aqueous solution of KI means 200 g of KI is present in 1000 g of solution.


Therefore,


Molality = 



= 1.506 m


∴ Molality of KI = 1.506 m


(b) Molarity is the concentration of a solution expressed as the number of moles of solute per litre of solution.


Given,


Density of the solution = 1.202 g/mL


Volume of 100 g solution = 



= 83.19 mL


Therefore, molarity = mol/ 83.19 × 10 - 3 L


= 1.45 M


∴ Molarity of KI = 1.45M


(c) Molar mass of KI  39 + 127 = 166 g/mol.


Moles of KI  = 0.12 mol


Moles of water =  = 4.44 mol


Therefore,


Mole fraction of KI = 



= 0.0263


∴ Mole fraction of KI = 0.0263




Intext Questions Pg-41
Question 1.

H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.


Answer:

According to Henry's law,


"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."


Stated as,


p = KHx


Where,


P = partial pressure of the solute above the solution


KH = Henry's constant


x = concentration of the solute in the solution


Given,


Solubility of H2S in water at STP is 0.195 m


We know,


At STP pressure p = 0.987 bar.


0.195 mol of H2S is dissolved in 1000g of water


Moles of water = 


= 55.56 g/mol


∴ the mole fraction of H2S = 



= 0.0035


According to Henry's law,


 p = KHx


 KH = 


 KH = 


 KH = 282 bar


∴ The Henry’s law constant is 282 bar



Question 2.

Henry’s law constant for CO2 in water is 1.67x108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atmCO2 pressure at 298 K.


Answer:

According to Henry's law,


"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."


Stated as,


p = KHx


Where,


P = partial pressure of the solute above the solution


KH = Henry's constant


x = concentration of the solute in the solution


Given,


KH = 1.67x108 Pa


PCO2 2.5 atm = 2.5 × 1.01325 × 105Pa


According to Henry's law,


 p = KHx




 x = 0.00152


In 500 ml of soda water there is 500 ml of water (neglecting soda)


Mole of water = 


= 27.78 mol


Now,


x = 


nCO2 = x  nH2O


= 0.00152 × 27.78


= 0.042 mol


Hence, quantity of CO2 in 500mL of soda water  0.042 × 44 = 1.848 g




Intext Questions Pg-47
Question 1.

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.


Answer:

Given,


PAo = 450 mm Hg


PBo = 700 mm Hg


ptotal = 600 mm of Hg


By using Rault's law,


ptotal = PA + PB


ptotal = PAoxA + PBoxB


ptotal = PAoxA + PBo( 1 - xA )


ptotal = (PAo- PBo)xA + PBo


600 = (450 - 700) xA + 700


-100 = -250 xA


xA = 0.4


∴ xB = 1 - xA


xB = 1 – 0.4


xB = 0.6


Now,


PA = PAoxA


PA = 450 × 0.4


PA = 180 mm of Hg


and


PB = PBox


PB = 700 × 0.6


PB = 420 mm of Hg


Composition in vapour phase is calculated by


Mole fraction of liquid, 



= 0.30


Mole fraction of liquid, 



= 0.70


Note: Alternate method to find the Mole fraction of liquid B is


= 1 - Mole fraction of liquid A


= 1 – 0.30


= 0.70




Intext Questions Pg-55
Question 1.

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.


Answer:

Given,


Vapour pressure of water, PIo = 23.8 mm of Hg


Weight of water, w1 = 850 g


Weight of urea, w2 = 50 g


Molecular weight of water, M1 = 18 g/mol


Molecular weight of urea, M2 = 60 g/mol


n1 =  = 47.22 mol


n2 =  = 0.83 mol


We have to calculate vapour pressure of water in the solution p1


By using Rault's therom,




(relative lowering)


 PI = 23.4 mm of Hg


Hence,


The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.



Question 2.

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.


Answer:

Given,


Mass of water, wl = 500 g


Boiling point of water = 99.63°C (at 750 mm Hg).


Molal elevation constant, Kb = 0.52 K kg/mol


Molar mass of sucrose (C12H22O11), M2 (11 × 12 + 22 × 1 + 11 × 16) = 342 g/mol


Elevation of boiling point ΔTb = (100 + 273) - (99.63 + 273) = 0.37 K


We know that,


ΔTb = 


0.37 = 


w2 = 


w2 = 121.67 g


Hence,


121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.



Question 3.

Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol-1.


Answer:

Given,


Mass of acetic acid, w1 = 75 g


Lowering of melting point, ΔTf = 1.5 K


Kf = 3.9 K kg/mol


Molar mass of ascorbic acid (C6H8O6), M2 6 × 12 + 8 × 1 + 6 × 16 = 176 g/mol[1]


We know that,


ΔTb = 


1.5 = 


w2 = 


= 5.08 g


Hence,


5.08 g of ascorbic acid is needed to be dissolved.



Question 4.

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.


Answer:

Given,


Volume of water, V = 450 mL = 0.45 L


Temperature, T  (37 + 273)K = 310 K


1.0 g of polymer of molar mass 185,000


Number of moles of polymer, n = 1 / 185,000 mol


We know that,


Osmotic pressure, π = 



= 30.98 Pa


= 31 Pa (approx)




Exercises
Question 1.

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.


Answer:

solution is a homogeneous mixture of two or more than two substances on molecular level whose composition can vary within certain limits.

The part or component of the mixture present in a lesser amount is called the SOLUTE and the one present in larger amount is called the SOLVENT. For eg- Small amount of salt [solute] dissolved in water [solvent].



There are nine types of solutions formed. They are:



Out of these nine types solution, solid in liquid, liquid in liquid & gas in liquid are very common. When the components of the solution are mixed, the resulting solution may exist in any of the three possible states of matter that is solid, liquid or gaseous.


They are:


[i] Gaseous solution: In such solutions solvent is gas. Since the solvent is gas, the solute can be solid, liquid or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.


[ii] Liquid solution: In such type of solutionsliquid acts as the solvent. The solute in these solutions may be gas, liquid, or solid.


[iii] Solid solutions: As the name suggests, in such solutions solid acts as the solvent. The solute in these solutions may be a gas, liquid or solid. For example, a solution of copper in gold is a solid solution.



Question 2.

Give an example of a solid solution in which the solute is a gas.


Answer:

As the name signifies, a solid solution is one in which solvent is solid.

So considering this aspect absorption of hydrogen over platinum or palladium is an example of such solution. Platinum or palladium is used as a catalyst in hydrogenation processes.



Question 3.

Define the following terms:

(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.


Answer:

(i) Mole fraction - The mole fraction of a particular component in a solution is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution.


Mathematically,


Mole Fraction of component


Mole Fraction is independent of temperature.


(ii) Molality - Molality of a solution is defined as the number of moles of solute dissolved per 1000g [1kg] of the solvent. It is represented by m.


Molality actually represents the concentration of solution in mol / kg.


Mathematically,


Molality 


It is represented by m.


(iii) Molarity- The number of moles of solute dissolved per litre of the solution at a particular temperature is called the molarity of the solution at that temperature.


Molarity actually represents the concentration of a solution in mol / L.


Mathematically,


Molarity 


(iv) Mass percentage - Mass percentage is defined as the mass of the solute in grams dissolved per 100g of the solution. It is also referred to as weight percentage [w/w].


For example, 10% [by mass] urea solution means that 10 g of urea are present in 100 g of solution, the solvent being only 100-10 = 90 g.


Mathematically, the mass percentage of a solute in a solution is given by:


Mass Percentage of Solute


Or


Mass Percentage of Solute 



Question 4.

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?


Answer:

Given:

Concentration of Nitric Acid, HNO3 = 68%


Density of solution, d = 1.504 g/ml


To find: Molarity, Mo


Formula:


Density, d


Molarity, Mo


Solution:


68% of Nitric acid by mass in aqueous solution means that 68g [[68 × 100]/100] of Nitric acid present in 100g of solution.


⇒ Molecular mass of Nitric Acid, HNO3 = [1 × 1] + [1 × 14] + [16 × 3]


= 63g


⇒ Number of moles of Nitric Acid = [68/63]


= 1.079 moles


⇒ Given Density, d = 1.504 g/ml


⇒ Volume, v = [100/1.504]


= 66.489 ml


⇒ Molarity, Mo = [1.079/66.489] × 1000


= 16.23 M


Therefore the molarity of the sample is 16.24 M.



Question 5.

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL–1, then what shall be the molarity of the solution?


Answer:

Given:

10% w/w solution means that if solution is of 100g then 10g of glucose is present in it [[10 × 100]/100] and the amount of water present in it is [100-10], Mw = 90g.


Density, d = 1.2g/ml


To find: Molality and Mole Fraction of Each component


Formula:


Density, d


Molality 


Mole Fraction of component 


Molarity, Mo


Solution:


Calculation of Molality:


⇒ Molecular mass of Glucose, [C6H12O6] = [6 × 12] + [12 × 1] + [6 × 16]


= 72 + 12 + 96


= 180g


⇒ Number of moles of Glucose, Ng = [10/180]


= 0.0556 moles


⇒ Molality 


⇒ Molality


= 0.6177 m


≈ 0.62 m


Calculation of Mole Fraction:


⇒ Molecular Mass of Water, [H2O] = [2 × 1] + [16 × 1]


= 2 + 16


= 18g


⇒ Number of moles of water, Nw = [90/18]


= 5 moles


⇒ Mole Fraction of Glucose, 



= 0.011


⇒ Mole Fraction of Water, 


= 1 – 0.011


= 0.989


Calculation of Molarity:


From density we can find out volume.


⇒ Volume, V = [100/1.2]


= 83.33 ml


⇒ Molarity 


Molarity, M = 0.667 M


≈ 0.67 M


Therefore the answers obtained in the numerical can be summarized in following table:




Question 6.

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?


Answer:
Molarity of HCl = 0.1 M Na2CO3 and NaHCO3 = 1 g
The weight of mixture of
The volume HCl required is 157.88 ml.

Question 7.

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.


Answer:

Let the first solution be solution A and second solution be solution B.


Given:


⇒ Mass of Solute in solution A, Ma = [25 × 300]/100


= 75g


⇒ Mass of Solute in solution B, Mb = [40 × 400]/100


= 160g


To find: Mass Percentage of Solute


Formula:


Mass Percentage of Solute


Solution:


⇒ Total mass of Solute in mixture, M = Ma + Mb


= [75 + 160]


= 235g


⇒ Total mass of solution = 300 + 400


= 700g


⇒ Mass Percentage of Solute 


= 33.5%


⇒ Percentage of Water in final solution = 100-33.5


= 66.5%


Therefore mass percentage of solute is 33.5% and that of water is 66.5%.



Question 8.

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution?


Answer:

Given:

Mass of ethylene glycol (C2H6O2) = 222.6 g


Mass of water = 200 g


Density, d = 1.072 g/ml


To find: Molality and Molarity of solution


Formula:


Molality 


Molarity, Mo


Density, d


Solution:


Calculation of Molality:


⇒ Molecular Mass of ethylene glycol (C2H6O2) = [12 × 2] + [6 × 1] + [16 × 2]


= 24 + 6 + 32


= 62 g


⇒ Number of moles of ethylene glycol (C2H6O2) = [222.6/62]


= 3.59 moles


⇒ Mass of water = 200 g


⇒ Molality 



= 17.95 m


Calculation of Molarity:


⇒ Total mass of solution = [222.6 + 200]


= 422.6 g


From density formula we can find out the volume required.


⇒ Volume of solution, V = [422.6 /1.072]


= 394.216 ml


⇒ Molarity, Mo


⇒ Molarity, Mo


⇒ Molarity = 9.1067


⇒ Molarity ≈ 9.11 M


Therefore the Molality and Molarity of the solution is as follows:


Molality = 17.95 m


Molarity = 9.11 M



Question 9.

A sample of drinking water was found to be severely contaminated with chloroform (CHCl3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):

(i) express this in percent by mass

(ii) determine the molality of chloroform in the water sample.


Answer:

Given:

Level of contamination = 15 ppm [by mass]


To find: Mass Percentage and Molality


Formula:


Molality 


Mass Percentage of Solute


Solution:


Calculation of Mass Percentage:


15 ppm means 15 parts of Chloroform in 106 parts of drinking water


⇒ Mass Percentage 



= 1.5 × 10-3


Calculation of Molality:


⇒ Molecular Mass of Chloroform, CHCl3 = [12] + [1] + [35.5 × 3]


= 119.5 g


⇒ Number of Moles of Chloroform = [15 / 119.5]


= 0.1255 moles


Molality 



= 1.255 × 10-4


Therefore the Mass Percentage is = 1.5 × 10-3 and the Molality of the solution is = 1.255 × 10-4 m.



Question 10.

What role does the molecular interaction play in a solution of alcohol and water?


Answer:

The lower members of alcohols are completely miscible [highly soluble] with water but the solubility decreases with increase in the molecular weight. The lower members of the alcohol group have the capability to form intermolecular hydrogen bonding with water molecules as alcohols are polar molecules in nature.


Alkyl groups are hydrophobic [prevents formation of hydrogen bonds with water] in nature. In lower alcohols, the alkyl group is small and the –OH group of alcohol is effective in making hydrogen bonds with water.


But with the increase in the size of alkyl group, the hydrophobic [water hating] nature of alkyl group dominates over the hydrophilic [water liking] nature of –OH group making the molecule less soluble in water. So solubility of alcohols decreases with increase in its molecular mass.


Since molecular interaction is weaker between higher alcohols and water, as a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily get liberated off. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.



Question 11.

Why do gases always tend to be less soluble in liquids as the temperature is raised?


Answer:

Whenever a gas is dissolved in a liquid, a small amount heat is liberated in the process. So dissolving a gas in liquid is overall an exothermic process.


So according to the LeChatelier principle, whenever the temperature is increased for a reaction which is exothermic in nature, the equilibrium shifts backwards and the reaction proceeds in backward direction that means the solution gets dissociated and will give off gas and hence solubility of gas decreases.


So with the increase in temperature, the solubility of the gases in liquids decreases.



Question 12.

State Henry’s law and mention some important applications?


Answer:

The solubility of a gas in water depends on following three parameters:

1. Nature of gas


2. Temperature


3. Pressure


The solubility decreases with increase in temperature. Temperature and pressure follow inverse proportionality. So solubility increases with increase with pressure. A quantitative relation between pressure and solubility of a gas in a solvent was given by W. Henry [1803]. This relationship is known as Henry’s law.


Statement:


Henry’s law can be expressed as follows.


At constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.


Mathematically,


Solubility ∝ Pressure of the gas


Some of the important applications of Henry’s law are as follows.


[i] Since the solubility of a gas in water increases with pressure, soft drink bottles are sealed under high pressure to accommodate more CO2 in the soft drink making the drink fizzier.


[ii] If deep sea divers [scuba divers] use air for respiration, they develop a medical condition known as bends with involves the blocking of capillaries. This is because air is mainly a mixture of oxygen and nitrogen. According to Henry’s Law, the solubility of gases increases with increase in pressure. When diver breathes air under high pressure in water, nitrogen dissolves in his blood. When the diver comes towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles in the blood. This is quite painful and dangerous to life. In order to avoid bends and toxic effects of dissolved nitrogen in blood, the tanks used by sea divers are filled with air diluted with helium [ 11.7% He + 56.2% N2 + 32.1% O2 ].


[iii] The pressure of oxygen in air decreases in going up the mountains. At very high altitudes the partial pressure of oxygen in air is much less than that at the ground level. Therefore, people living at high altitude or climbers have low concentrations of oxygen in the blood and tissues. This leads to weakness and loss in the clarity of thinking. These symptoms create a condition known as anoxia.


[iv] Henry’s law is also related to biology as it explains the supply of inhaled oxygen to tissues. When air is inhaled, it combines with haemoglobin [oxygen carrier of RBCs] of the blood in lungs to form oxyhaemoglobin because in lungs the partial pressure of oxygen is high. Partial pressure of oxygen is low in tissues. Hence, oxygen is released from oxyhaemoglobin and is utilised by the cells to carry out their tasks.



Question 13.

The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?


Answer:

Given:

Mass of ethane in solution A = 6.56 × 10–3 g


Partial pressure of solution A = 1 bar


Mass of ethane in solution B = 5.00 × 10–2 g


To find: Partial Pressure of gas


Formula:


By Henry’s law:


Mass of dissolved gas M = k × P


Where


k = proportionality constant


P = Partial Pressure


Solution:


⇒ M1 = k × P1 ……………… [1]


⇒ M2 = k × P2 …………….. [2]


Dividing the [2] by [1], we get





P2 = 7.62 bar


Therefore the partial pressure of the gas is 7.62 bar.



Question 14.

What is meant by positive and negative deviations from Raoult's law and how is the sign of Δmix related to positive and negative deviations from Raoult's law?


Answer:

Raoult’s law states that at a given temperature, the vapour pressure of a solution containing non volatile solute is directly proportional the mole fraction of the solvent.

Non ideal solutions show positive and negative deviations from ideal behaviour.


Non ideal solutions showing positive deviations from Raoult’s law-


A solution is said to show positive deviation from Raoult’s Law when at any composition, its vapour pressure is higher than that given by Raoult’s Law.


The positive deviation is shown by those liquid pairs in which the A-B molecular interaction forces are weaker than the corresponding A-A or B-B molecular interaction forces. When the A-B molecular interaction forces are weaker, then molecules of liquid A find it easier to escape as compared to pure solution thereby increasing the vapour pressure of solution.


As a result, each component of solution has a partial vapour pressure greater than expected on the basis of Raoult’s law. This is called positive deviations from Raoult’s law, that is PA>PAoXA and PB>PBoXB


mixH [Change in Enthalpy] is positive because energy is required to break A-A & B-B attractive forces. Hence it is an endothermic process.


Examples of liquid Pairs:


1. Water + methanol


2. Water + ethanol


3. Acetone + benzene


4. Carbon tetrachloride + benzene


The graph of vapour pressure and mole fraction for liquids showing positive deviation from Raoult’s Law is shown below:


From the diagram it is clear that the composition curve of such a solution lies above the composition curve obtained on basis of Raoult’s Law.



Non ideal solutions showing Negative deviations from Raoult’s law –


When the vapour pressure of the solution is lower than that calculated on the basis of Raoult’s Law, the solution is said to show negative deviation from Raoult’s Law. Even the vapour pressure of individual components show negative deviation from Raoult’s Law


The negative deviation is shown by those liquid pairs in which the A-B molecular interaction forces are stronger than the corresponding A-A or B-B molecular interaction forces. When the A-B molecular interaction forces are stronger, then molecules of liquid A find it easier to escape from pure solution as compared to the mixture thereby decreasing the vapour pressure of solution.


Consequently, each component of solution has a partial vapour pressure less than expected on the basis of Raoult’s law. This is called negative deviations from Raoult’s law, i.e. PA<PAoXA & PBoXB.


Examples of liquid pairs:


1. Water + hydrochloric acid


2. Water + nitric acid


3. Chloroform + benzene


4. Acetone + aniline


mixH is negative because energy is released due to increase in attractive forces. Hence it is an exothermic process.


The graph of vapour pressure and mole fraction for liquids showing negative deviation from Raoult’s Law is shown below:




Question 15.

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?


Answer:

Given:

2% non-volatile solute in an aqueous solution means that if the solution is of mass 100 g then the mass of solute in the solution is,


Mass of solute = [2 × 100]/100


= 2 g


∴ Mass of Solvent = 100-2


= 98 g


Vapour pressure of solution at normal boiling point, P1 = 1.004 bar


Vapour pressure of pure water at normal boiling point, P1° = 1 atm = 1.013 bar


To find: Molar mass of solute


Formula:


From Raoult’s Law, we have,


 ………….. [1]


Now we know that number of moles of solute is given by the following relationship,


⇒ Number of moles 


Using the above relationship the equation [1] can be modified as follows:



Where


W1 = mass of solvent


W2 = mass of solute


M1 = molecular mass of solvent


M2 = molecular mass of solute


Solution:




M2 = 


M2 = 41.3469 g


∴ M2 ≈ 41.35 g


Therefore the molar mass of the solute is 41.35 g.



Question 16.

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?


Answer:

Given: Temperature = 373k

Vapour pressure of pure heptane (p01) = 105.2kpa and that of octane (p02) = 46.8kpa


Mass of heptane = 26 g


Mass of octane = 35 g


Molecular weight of heptane = C7H16 = 12 × 7 + 1 × 16 = 100 gmol-1


Molecular weight of octane = C8H18 = 114 gmol-1


Moles of heptane, n1 = given mass /molecular weight = 26/100


⇒ n1 = 0.26mol


Moles of octane, n2 = given mass /molecular weight = 35/114


⇒ n2 = 0.307mol


∴ Mole fraction of heptane, 


⇒ 


⇒ χ1 = 0.456


Now, Mole fraction of octane,


⇒ 


⇒ χ2 = 0.544


∴ Partial pressure of heptane, p1 = χ1 × p01


⇒ p1 = = 0.456 × 105.2 = 47.97kpa


∴ Partial pressure of octane, p1 = χ2 × p02


⇒ p2 = = 0.544 × 46.8 = 25.46 kpa


∴ Total pressure exerted by solution = p1 + p2


= 47.97 + 25.46


= 73.43kpa



Question 17.

The vapour pressure of water is 12.3kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.


Answer:

Given: 1 molal solution means 1 mole of solute present in 1000g of water solvent)

Molecular weight of water = H2O = 1 × 2 + 16 = 18g/mol


No. of moles of water, n = given mass /molecular weight


⇒ n = 1000/18 = 55.56 gmol-1


Mole fraction of solute in solution,x2 = moles of solute/(moles of solute + moles of water)


⇒ x2 = 1/(1 + 55.56)


⇒ x2 = 0.0177


Given vapour pressure of pure water at 300k is 12.3 kpa


Apply the formula:



⇒ 


⇒ P1 = 12.0823kpa


which is the vapour pressure of the solution.



Question 18.

Calculate the mass of a non-volatile solute (molar mass 40 g) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.


Answer:

Given:


Molar mass of non-volatile solute = 40g


Let no. of moles of solute be n.


Mass of octane = 114g


Molar mass of octane(C8H18) = 12 × 8 + 1 × 18 = 114g/mol


Moles of octane = given mass/molar mass


⇒ n = 114/114 moles


⇒ n = 1 mole


Molar fraction of solute,


⇒ x2 = n/n + 1


Let the vapour pressure of original solvent(without solute) be p1


Accordingly after addition of solute vapour pressure of solution reduces to 80% i.e. 0.8 p1 = p1


Applying the formula:



⇒ 


⇒ n/n + 1 = 0.2


⇒ 0.2n + 0.2 = n


⇒ n = 0.25 moles


Hence, mass of solute is:


moles = given mass/molar mass


⇒ 0.25moles = mass/40g


⇒ mass = 10g.



Question 19.

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

i. molar mass of the solute

ii. vapour pressure of water at 298 K.


Answer:

given: mass of solute = 30g

Let the molar mass of solute be x g and vapour pressure of pure water at 298k be P1


Mass of water(solvent) = 90g


Molar mass of water = H2O = 1 × 2 + 16 = 18g


Moles of water = mass of water/molar mass


⇒ n = 90/18 moles


⇒ n = 5moles


Molar fraction of solute,


⇒ 


⇒ 


Vapour pressure of solution(p1) = 2.8kpa


Applying the formula:


⇒ 


⇒ 


⇒ 


⇒  -(1)


According to second condition when we add 18g of water to solution vapour pressure becomes 2.9kpa


Moles of water = mass/molar mass


⇒ n = 90 + 18/18


⇒ n = 6moles


Molar fraction of solute,


⇒ 


⇒ 


Applying the formula:


⇒ 


⇒ 


⇒ 


⇒  -(2)


Solving 1 and 2:


Dividing (2) by (1) we get


⇒ 


⇒ 


0.8631(30 + 6x) = 30 + 5x


25.9 + 5.18x = 30 + 5x


0.18x = 4.1


X = 22.78g


Substituting value of x in 1 we get


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


i. molar mass of the solute = 22.78g


ii. vapour pressure of water at 298 K = 3.537kpa



Question 20.

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.


Answer:

In cane sugar:

ΔTf= (273.15 - 271) K = 2.15 K

Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol- 1

5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g = 95 g of water.

Now, number of moles of cane sugar =5/342 mol

= 0.0146 mol

So, molality of the solution,m=0.0146mol/0.095kg

= 0.1537 kg mol- 1

Relating,

ΔTf= Kf× m

⇒ Kf= ΔTf / m

⇒ 2.15K / 0.1537 kg mol-1

= 13.99 K kg mol-1

Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol- 1

5% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water.

∴ No of moles of glucose = 5/180 mol

= 0.0278 mol

So, molality of the solution,m=0.0278 mol / 0.095 kg

= 0.2926 mol kg- 1

Applying the relation,

ΔTf = Kf × m

= 13.99 K kg mol- 1× 0.2926 mol kg- 1

= 4.09 K

Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.


Question 21.

Two elements A and B form compounds having formula AB2 and AB4 . When dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1K kg mol-1. Calculate atomic masses of A and B.


Answer:

let the molar masses of AB2 and AB4 be x and y respectively.

Molar mass of benzene(C6H6) = 12 × 6 + 1 × 6 = 78g/mol


Moles of benzene = mass/molar mass = 20/78


n = 0.256mol


⇒ ΔTf = 2.3k


Kf = 5.1K kg mol-1


For AB2


Applying the formula: ΔTf = Kf × M


Where


ΔTf = depression in freezing point


Kf = molal depression constant


M = molality of solution


⇒ 2.3 = 5.1 × M1


⇒ M1 = 0.451mol/kg


For AB4


Applying the formula: ΔTf = Kf × M


Where


ΔTf = depression in freezing point


Kf = molal depression constant


M = molality of solution


⇒ 1.3 = 5.1 × M2


⇒ M2 = 0.255mol/kg


M1 = moles of solute/mass of solvent(in kg)


⇒ 


⇒ X = 110.86g


M2 = moles of solute/mass of solvent(in kg)


⇒ 


⇒ y = 196.1g


atomic mass of A be a and that of B be b g respectively.


So, AB2 : a + 2b = 110.86


AB4 : a + 4b = 196.1


⇒ a = 25.59g


⇒ b = 42.64g



Question 22.

At 300 K, 36 g of glucose present in a litre of its solution has a osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what would be its concentration?


Answer:

Here,

T = 300 K


π = 1.52 bar


R = 0.083 bar L


Applying the relation, π = CRT


Where


π = osmotic pressure of solution


C = concentration of solution


R = universal gas constant


T = temperature


⇒ 


⇒ C = 0.061mol/L


Concentration of the solution is 0.061mol/L



Question 23.

Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I2 and CCl4

(iii) NaClO4and water

(iv) methanol and acetone

(v) acetonitrile(CH3CN) and acetone(C3H6O)


Answer:

(i) Both the compounds are non-polar and they do not attract each other because they do not form any polar ions. Vanderwaals forces of attraction will be dominant in between them as vanderwaals forces of attraction are not a result of any chemical or electronic bond.

(ii) now here both the compounds are non-polar because in I2 both the atoms are same so they have same electronegativity and hence there will be no displacement of electron cloud, it will be in the centre. In case of CCl4 molecule, it has tetrahedral shape so two Cl atoms will cancel the attraction effect from two opposite Cl atoms, hence molecule as a whole is non polar. Therefore they will also have vanderwaals forces of attraction.


(iii) NaClO4 is ionic in nature as Na, Cl and O all have different electronegativity and their shape is also not symmetric. So molecular will be ionic in nature and we know that water is polar because O will attract the electron cloud towards it(more electronegative) hence there will be formation of dipole(two oppositely charged ions separated by a short distance). Therefore there will be ion-dipole interaction between them.


(iv) methanol and acetone are both polar molecules because of the presence of electron withdrawing O atom in methanol and ketone group in acetone. So they will have dipole-dipole interaction.


(v) acetonitrile is polar compound due to presence of electronegative N atom and acetone due to ketone group. So, dipole-dipole interaction.



Question 24.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3CN, CH3OH.


Answer:

Now n-octane is non-polar solvent due to long chain saturated structure. We know that “like dissolves like” so a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.

So cyclohexane is non-polar due to symmetric structure. KCl is ionic in nature as it will dissociate into K + and Cl- ions. CH3CN is polar as mentioned above and CH3OH is also polar in nature.


The order of increasing polarity is:


Cyclohexane < CH3CN < CH3OH < KCl (O is more electronegative than N)


Therefore, the order of increasing solubility is:


KCl < CH3OH < CH3CN < Cyclohexane



Question 25.

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i) phenol

(ii) toluene

(iii) formic acid

(iv) ethylene glycol

(v) chloroform

(vi) pentanol


Answer:

Water is a polar compound (due to electronegativity difference between O and H) . We know that “like dissolves like”. So, a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.

(i) Phenol has the polar group -OH and non-polar group –C6H5 and it can not form H bonding with water(presence of bulky non-polar group) . Thus, phenol is partially soluble in water.


(ii) Toluene has no polar groups. Thus, toluene is insoluble in water.


(iii) Formic acid (HCOOH) has the polar group -OH and can form H-bond with water. Thus, formic acid is highly soluble in water.


(iv) Ethylene glycol(OH-CH2-CH2-OH) has polar -OH group and can form H-bond with water. Thus, it is highly soluble in water.


(v) Chloroform is partly soluble as CHCl3 is polar in nature due to high electronegativity of Cl atoms, there will be production of partial + charge on H atom so it can form H bonding with water but it is also surrounded by 3 Cl atoms, so partly soluble.


(vi) Pentanol(C5H11OH) has polar -OH group, but it also contains a very bulky non-polar group –C5H11. Thus, pentanol is partially soluble in water.



Question 26.

If the density of some lake water is 1.25g/ml and contains 92 g of Na+ ions per kg of water, calculate the molality of Na+ ions in the lake.


Answer:

Mass of ions = 92g


Molar mass of ions = Na+ = 23g(neglect the mass lost due to absence of a electron)


Moles of ions = mass of ions/molar mass


⇒ n = 92/23 moles


⇒ n = 4moles


Molality of solution = moles of solute/mass of solvent(in kg)


Molality = 4/1 = 4M



Question 27.

If the solubility product of CuS is 6 × 10-16, calculate the maximum molarity of CuS in aqueous solution.


Answer:

The Solubility product of CuS (ksp) = 6 × 10-16

CuS → Cu + + + S2-


Let the s be solubility of CuS in mol/L


Ksp = [ Cu + + ][S2]


Ksp = solubility product


6 × 10-16 = s × s = s2


⇒ S = 2.45 × 10-8 mol/L


Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10-8 mol/L



Question 28.

Calculate the mass percentage of aspirin(C9H8O4) in acetonitrile(CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.


Answer:

Total mass of solution = 6.5g + 450g = 456.5g

Therefore mass percentage of aspirin in solution = (mass of aspirine/total mass of solution) = 6.5/456.5


⇒ mass % of aspirine = 1.424%



Question 29.

Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10-3 m aqueous solution required for the above dose.


Answer:

molar mass of Nalorphene = 311g/mol

Now 1000g of solution contains 1.5 × 10-3 moles of Nalorphene(Molality of solution = moles of solute/mass of solvent(in kg))


⇒ 1.5 × 10-3 moles of Nalorphene = 1.5 × 10-3 × 311 = 0.4665g of Nalorphene


Therefore, total mass of the solution = (1000 + 0.4665) g


⇒ total mass = 1000.4665 g


This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.


Therefore, mass of the solution containing 1.5 mg of nalorphene is:



⇒ mass of solution containing required ions = 3.22 g


Hence, the mass of aqueous solution required is 3.22 g.



Question 30.

Calculate the amount of benzoic acid required for preparing 250mL of 0.15M solution in methanol.


Answer:

Molarity = moles of the solute / volume of solution

Using the above formula we get,

0.15 = (mole of benzoic acid)/250 x 1000

Or

moles of benzoic acid = (0.15 x 250)/1000

= 0.0375 mol of benzoic acid

Molecular mass of benzoic acid with the formula C6H5COOH

= 7 × 12 + 6 × 1 + 2 × 16

= 122 g/mol

Hence, amount of benzoic acid = 0.0375 x 122 = 4.575 g


Question 31.

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.


Answer:

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the pattern,


Acetic acid< trichloroacetic acid< trifluoroacetic acid.


This is because fluorine is more electronegative than chlorine. So, trifluoracetic acid is a stronger acid in comparison to trichloroacetic acid and acetic acid. And also, acetic acid is the weakest of all.


Explanation: Stronger acid produces more number of ions, therefore it has more ΔTf(depression in freezing point), hence lower freezing point. As the acidic strength increases, the acid gets more and more ionised.


Trifluoracetic acid ionizes to the largest extent. Hence, in this case, trifluoracetic acid being the strongest acid produces more number of ions(extent of ionisation and concentration of ions are more), high ΔTf(depression in freezing point)and lower freezing point and vice versa.



Question 32.

Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1.


Answer:

Given: Mass of CH3CH2CHClCOOH = 10 g


Mass of water = 250g


Ka = 1.4 × 10–3,


Kf = 1.86 K kg mol–1


Molar mass of CH3CH2CHClCOOH = 12 + 3 + 12 + 2 + 12 + 1 + 35.5 + 2 + 16 + 16 + 1


= 122.5 g mol–1


Number of moles of solute = 


→ No. of moles = 


∴ No. of moles = mol


Now, Molality is given as,




M = 0.3264 kg/mol


CH3CH2CHClCOOH CH3CH2CHClCOO- + H +



Total moles at equilibrium = (1-α) + 2 α


= 1 + α


In order to find out the depression in freezing point, ff values of i(vant Hoff’s factor) and α(degree of dissociation) are to be found out.


To find out degree of dissociation, α


→ 


Here, the value of α is negligible as compared to 1, and hence 1- = 1, giving,





∴ 


To find out Vant Hoff’s factor,


→ 




∴ 


Now, to find out the depression in freezing point,


ff




∴ 


Thus, the depression in freezing point isf= 0.650C



Question 33.

19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the vant Hoff factor and dissociation constant of fluoroacetic acid.


Answer:

Given- w1 = 500g


W2 = 19.5g


Kf = 1.86 K kg mol-1


Molar mass of CH2FCOOH = 12 + 2 + 19 + 12 + 16 + 16 + 1


= 78 g mol-1


The depression in freezing point is calculated by,


→ ff (where, m is the molality)





∴ f (calculated) = 0.93


To find out the vant Hoff’s factor, we use the formula,


→ 



∴ 


CH2FCOOH → CH2FCOO- + H +


To find out the degree of dissociation α, we use


→ 


∴ 


With degree of dissociation, known to us, we can easily calculate the dissociation constant, Ka using the formula,




∴ 


Thus, the vant Hoff’s factor is 1.07 an the dissociation constant is 2.634x10-3



Question 34.

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.


Answer:

Given- Vapour pressure of water,  = 17.535 mm Hg


 = 25 g of glucose


 = 450g of water


Molar mass of water, H2O = 1 + 1 + 16 = 18 g mol-1


Molar mass of glucose, C6H12O6 = (12×6) + (1×12) + (16×6) = 180 g mol-1


Using Raoult’s law for solution of non-volatile solute,


 → Equation 1


where  is the mole fraction of the solute




Substituting the value of  in equation 1, we get,


→ 








Thus, the vapour pressure of water at 293 K at the given conditions is 17.437 mm Hg



Question 35.

Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.


Answer:

Given-


Henry’s law constant KH = 4.27X 105 mm Hg,


p = 760mm Hg,


Using Henry’s law,




∴ 


Using the formula of lowering vapour pressure,


 → Equation 1


Since, <<<, the Equation 1 is rewritten as,



where  and  = molarity of benzene





∴ 


Thus, the solubility of methane in benzene is 0.023 moles



Question 36.

100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.


Answer:

Given-


Mass of liquid A, WA = 100g, Molar mass, MA = 140 g mol-1


Mass of liquid B, WB = 1000 g, Molar mass, MB = 180 g mol-1


Using the formula below calculate the no. of moles in liquid A and B.


→ 


Number of moles of liquid A, MA = 100/140 = 0.714 mol-1


Number of moles of liquid B, MB = 1000/ 180 = 5.556 mol-1


Using the formula,


,


we calculate the mole fraction of liquids A and B.


→ Mole fraction of A, 


∴ 


→ Mole fraction of B, 



Vapour pressure of pure liquid B, PoB = 500 torr (given)


According to Henry`s law,


→ 




Having given the total vapour pressure of the solution, Ptotal = 475 torr,


→ 





Using Henry`s law again to get the vapour pressure of pure liquid A,





Thus, the vapour pressure of pure liquid A = 280.7 torr and


vapour pressure of liquid A in the solution = 32 torr



Question 37.

Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal, pchloroform, and pacetoneas a function of xacetone. The experimental data observed for different compositions of mixture is:


Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.


Answer:

The Ptotal for the values given in the graph is found out and plotted in the graph.




It can be observed from the graph that the plot for the p total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.



Question 38.

Benzene and naphthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.


Answer:

Given-


of benzene = 50.71 mm Hg


 of naphthalene = 32.06 mm Hg


Using the formula of lowering vapour pressure to calculate mole fraction of benzene and naphthalene in solution,





Thus, 



Using formula for vapour pressure,







Mole fraction of benzene in vapour phase is calculated by,





Thus, mole fraction of benzene in vapour phase is 0.6744



Question 39.

The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.


Answer:

Given-


KH for O2 = 3.30 × 107 mm Hg,


KH for N2 = 6.51 × 107 mm Hg


Percentage of oxygen (O2) = 20 %


Percentage of nitrogen (N2) = 79%


Total pressure = 10 atm


Using Henry’s law,




where, p is the partial pressure of gas in the solution and KH is Henry’s constant.


Now, to determine the mole fraction of oxygen in solution, , we use




To determine the mole fraction of nitrogen in solution, ,




Thus, themole fraction of oxygen in solution, xoxy = 4.61x10-5


and themole fraction of nitrogen in solution, xnit is 9.22x10-5



Question 40.

Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.


Answer:

Given-


Vant Hoff’s factor, i = 2.47


osmotic pressure,  = 0.75 atm


Volume of solution = 2.5L.


To determine the amount of CaCl2, we use vant Hoff’s equation for dilute


solutions, given as,



where, n is the number of moles of solute, R is solution constant which is equal


to the gas constant and T is the absolute temperature.




where, the molecular mass of CaCl2 is 111, t = 300 K (270C + 273 K) and R = 0.082,




Hence, the amount ofCaCl2 dissolved is 3.425g



Question 41.

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated.


Answer:

Given-


Mass of K2SO4, w = 25 mg = g,


Molar mass of K2SO4 = (39×2) + (32×1) + (16×4) = 174 g mol-1


Volume V = 2 liter


T = 250C + 273 = 298 K (add 273 to convert in Kelvin)


The reaction of dissociation of K2SO4 is written as,


K2SO4 → 2K + + SO42-


Number if ions produced = 2 + 1 = 3, hence vant Hoff’s factor, i = 3


Here, we use vant Hoff’s equation for dilute solutions, given as,


→ 


where, n is the number of moles of solute, R is solution constant which is equal to the gas constant(0.082) and T is the absolute temperature (298 K).






Hence, the osmotic pressure of a solution is 5.27x10-3atm